这是我的节点网格:
我正在使用 A* 寻路算法在其上移动一个对象。它通常可以正常工作,但有时会出现错误:
- 从 3 移动到 1 时,它正确地通过 2。但是,当从 1 移动到 3 时,它通过 4。
- 在 3 和 5 之间移动时,它会在任一方向上通过 4,而不是通过 6 的较短路径
有什么问题?这是我的代码(AS3):
public static function getPath(from:Point, to:Point, grid:NodeGrid):PointLine {
// get target node
var target:NodeGridNode = grid.getClosestNodeObj(to.x, to.y);
var backtrace:Map = new Map();
var openList:LinkedSet = new LinkedSet();
var closedList:LinkedSet = new LinkedSet();
// begin with first node
openList.add(grid.getClosestNodeObj(from.x, from.y));
// start A*
var curNode:NodeGridNode;
while (openList.size != 0) {
// pick a new current node
if (openList.size == 1) {
curNode = NodeGridNode(openList.first);
}
else {
// find cheapest node in open list
var minScore:Number = Number.MAX_VALUE;
var minNext:NodeGridNode;
openList.iterate(function(next:NodeGridNode, i:int):int {
var score:Number = curNode.distanceTo(next) + next.distanceTo(target);
if (score < minScore) {
minScore = score;
minNext = next;
return LinkedSet.BREAK;
}
return 0;
});
curNode = minNext;
}
// have not reached
if (curNode == target) break;
else {
// move to closed
openList.remove(curNode);
closedList.add(curNode);
// put connected nodes on open list
for each (var adjNode:NodeGridNode in curNode.connects) {
if (!openList.contains(adjNode) && !closedList.contains(adjNode)) {
openList.add(adjNode);
backtrace.put(adjNode, curNode);
}
}
}
}
// make path
var pathPoints:Vector.<Point> = new Vector.<Point>();
pathPoints.push(to);
while(curNode != null) {
pathPoints.unshift(curNode.location);
curNode = backtrace.read(curNode);
}
pathPoints.unshift(from);
return new PointLine(pathPoints);
}
NodeGridNode::distanceTo()
public function distanceTo(o:NodeGridNode):Number {
var dx:Number = location.x - o.location.x;
var dy:Number = location.y - o.location.y;
return Math.sqrt(dx*dx + dy*dy);
}